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**Unformatted text preview: **MIT OpenCourseWare http://ocw.mit.edu 14.30 Introduction to Statistical Methods in Economics Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms . n N 1 , 4 under H 25 A 14.30 Introduction to Statistical Methods in Economics Lecture Notes 21 Konrad Menzel May 5, 2009 Constructing Hypothesis Tests If X i has support S X , then the sample X = ( X 1 ,...,X n ) has support S n The critical region of a test is a region C X ⊂ S n of the support of the sample for which we reject the null. X . X The following example illustrates most of the important issues in a standard setting, so you should look at this carefully and make sure that you know how to apply the same steps to similar problems. Example 1 Suppose X 1 ,...,X n are i.i.d. with X i ∼ N ( µ, 4) , and we are interested in testing H 0 : µ = 0 against H A : µ = 1 . Let’s first look at the case n = 2 : X 1 X 2 A o Reject : C x Don't reject 1 1 k k We could design a test which rejects for values of X 1 + X 2 which are ”too large” to be compatible with µ = 0 . We can also represent this rejection region on a line: This representation is much easier to use if n is large, so it’s hard to visualize the rejection region in terms of X 1 ,...,X n directly. However, by condensing the picture from n to a single dimension we may loose the ability of specifying really odd-shaped critical regions, but typically those won’t be interesting for practical purposes anyway. ¯ So in this example, we will base our testing procedure on a test statistic T n ( X 1 ,...,X n ) = X n and reject for large values of T n . How do we choose k ? - we’ll have to trade off the two types of error. Suppose now that n = 25 , and since X i ∼ N ( µ, 4) , 4 N , 25 T n := X ¯ ∼ under H 1 Image by MIT OpenCourseWare. 1 1 n = 4 n = 25 n = 100 α β 1 0 k 2 X 1 + X 2 1 0 X n 4 k Don't reject Don't reject C x : reject C x : reject Now we can calculate the probabilities of type I and type II error k − 0 k ¯ α = P ( X > k | µ = 0) = 1 − Φ = Φ − 2 / 5 2 / 5 k − 1 ¯ β = P ( X ≤ k | µ = 1) = Φ 2 / 5 Therefore, fixing any one of α,β,k determines the other two, and that choice involves a specific tradeoff between the probability of type I and type II error - if we increase k , the significance level α goes down, but so does power 1 − β . Specifically, if we choose k = 5 3 , α ≈ 6 . 7% , and β ≈ 15 . 87% . For different sample sizes, we can graph the trade-off between the probability of type I and type II error through the choice of k as follows: A low value of k would give high power, but also a high significance level, so that increasing k would 2 move us to the left along the frontier....

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